12 Discrete Probability Distributions: Random variables, Expected Value and Variance:

Dr. Nidhi Handa

1. Concept of a random variable
2. Discrete random variable
3. Continuous random variable
4. Their  Probabilities ,Expected value,Mean & Variance
5. Related Problems
6.  Summary

CONCEPT OF A RANDOM VARIABLE:

Any given statistical experiment has more than one outcome. It is impossible to predict which of those outcomes will occur if that experiment is performed. Consequently decisions are made under certain conditions for example a horse rider does not know in advance whether or not he is going to win the race. If he knows that he is not going to win he will definitely not participate,. It is uncertainty about winning (on positive side) that makes him play. If, outcomes and probability for a statistical statement are known, we can find what will happen on average if that experiment is performed many times. For horse rider example, we can find out what a rider can expect to win (or lose) on average if he continues participating again and again.

Here following table gives the frequency and relative frequency distributions of the number of cable connections owned by all 2000 families in a small city.

Table— frequency and relative frequency distributions of the the number of cable connections owned by families

TABLE-1.

Number of cable connections	Frequency	Relative frequency
0	30	30/2000 =0.15
1	470	170/2000 = .235
2	850	850/2000 = .425
3	490	490/200 =.245
4	160	160/2000 = .080
	N = 2000	Sum = 1.000

Suppose one family is randomly selected from this population. The act of randomly selecting a family is called a random or chance experiment. Let, x denote the number of connection owned by selected families. Then x can assume any of the five possible values (0,1,2,3 and 4). Listed in the first column of above table. The valued assumed by x depends on which family is selected. Thus, this value depends on the outcome of a random experiment. Consequently x is called a random variable or chance variable. In general, a random variable is denoted by x or y.

A RANDOM VARIAQBLE IS A VARIABLE WHOSE VALUE IS DETERMINED BY OUTCOME OF RANDOM EXPERIMENT

If we toss two coins together, the sample space

S = {HH, HT, TH, TT}

Consist of four points. The probability of each outcome is ¼. Let the number of appearance of heads be dented by a variable X, which can assume values 0, 1 and 2

Now

p (x = 0) = probability of 0 head = {TT} = ¼

p (x = 1) = probability of 1 head = {HT, TH} = p (x = 2) = probability of 2 Head = {HH} = ¼

This can be arranged in the following tabular form :-

No. of Heads	Probabilities
0	¼
1	½
2	¼
Total	1.0

Such a distribution is called probability distribution and the variable x whose value depends on chance is called random variable

Consider the sample space giving detailed description of each possible outcome when three electronic components are tested. This can be written as

S = {NNN,  NND,  DNN, DND,  NDN, NDD, DDN, DDD}

When three electronic components are tested, x can assume values 0, 1, 2 & 3

Now p (x = 0) = p {NNN} = 1/8 , p {x = 1} = p {NND, NDN, DND} = 3/8 p(x = 2) = p{NDD, DND, DDN} = 3/8, p (x = 3) = p(DDD)=1/8.

The sum of all probabilities is one, the variable x is a random variable.

Definition : A random variable is a function defined on a sample space which associates a real number with each element of the sample space.

A random variable can be discrete or continuous.

If random variable takes on integer values such as 0, 1, 2, …… then it is called discrete

random variables.

In other words, a random variable that assumes countable values is called discrete random variable.The number of cars sold at a dealership during a given month.The number of shoe pairs a person owns etc. are examples of discrete random variable.

CONTINUOUS RANDOM VARIABLE

A random variable that can assume any value contained in one or more intervals is called continuous random variable. A random variable whose values are not countable is a continuous random variable. A continuous random variable can assume any value over an interval or intervals.

A RANDOM VARIABLE THAT CAN ASSUME ANY VALUE CONTAINED IN ONE OR MORE INTERVALS IS CALLED A CONTINUOUS RANDOM VARIABLES.

Because the number of values contained in any interval is infinite, the possible number of values that a continuous random variable can assume is also infinite. Moreover we cannot count these values. Consider the life of a electricity bulb. We can measure it precisely as we want. For example, the life of this bulb may be 40 hours or 40.25 hours or 40.247 hours. Assume that the maximum life of such bulb is 200 hours. Let x denote the life of a randomly selected bulbs of this kind. Then x can assume any value in the interval 0 to 200. Consequently, x is a continuous random variable. As shown below, every point on the line representing the interval 0 to 200 gives a possible value of x.

0___________________________________________200

(every point on this line represents a possible value of x that denotes the

life of a electricity bulbs. There are an infinite number of points on this line. The values represented by points on this line are uncountable)

The amount of rainfall on a rainy day or in a rainy season, the height and weight or individuals are examples of continuous random variable.

Probability Distribution Of A DISCRETE Random Variable

Let x be a discrete random variable .The probability distribution of x describes how the probabilities are distributed over all possible values of x.

The probability distribution of a discrete random variable lists all the possible values that the random variable can assume and their corresponding probabilities.

EXAMPLE 5-1 Reconsidering the frequency and relative frequency distributions of the number of the cable connections owned by the families in the Table 1. that table is reproduced as Table 2 on the next page. Let x be the number of vehicles owned by a randomly selected family. Write the probability distribution of x.

Number of cable connections	Frequency	Relative frequency
0	30	0.15

1	470	0.235
2	850	0.425
3	490	0. 245
4	160	0 .080
	N = 2000	Sum = 1.000

Solution Since the relative frequencies obtained from an experiment or a sample can be used as an approximate probabilities. However, when the relative frequencies are known for the population, they give the actual (theoretical) probabilities of outcomes. Using the relative frequencies of Table 2, we can write the probability distribution of the discrete random variable in the Table3.

TABLE 3

Number of cable connections	Probability p(x)
0	0.15
1	0.235
2	0.425
3	0. 245
4	0 .080
	∑  ( ) = 1

The probability distribution of a discrete random variable possesses the following two characteristics

1. The probability assigned to each value of a random variable x lies in the range 0 to 1 that is , 0≤ p(X) ≤ 1 for each x.
2. The sum of the probabilities assigned to all possible values of x is equal to 1.0 ,that is ,∑  ( )= 1. ( Note that if all the probabilities are rounded, the sum may not be exactly 1.0

TWO CHARACTERISTICS OF A PROBABLITY DISTRIBUTION

The probability distribution of a discrete random variable possesses the following two characteristics.

1. 0≤ P (x) ≤ 1for each value of x
2. 2. ∑ P(x) = 1

These two characteristics are also called the two conditions that a probability distribution must satisfy. Notice that in Table 3, each probability listed in the column labeled P(x) is between 0 and 1. Also, ∑ P(x) = 1.0. Because both conditions are satisfied, Table 3 represents the probability distribution of x.

From Table 3, the probability for any value of x can be read. For example, the probability that a randomly selected family from this town owns two cable connection is .425. The probability is written as

P(x = 2)  = .425

The probability that the selected family owns more than two connection is given by the sum of the probabilities of three and four cable connections, respectively. This probability is

.245 + .080 = .325. This probability is written as P(x > 2) = P(x = 3) + P(X =4) = .245 + .080 = .325

The probability distribution of a discrete random variable can be presented in the form of a mathematical formula, a table, or a graph.

PROBABILITY DISTRIBUTION

A frequency distribution is a listing of the observed frequencies of all the outcomes of an experiment that actually occurred when the experiment is done.

A probability distribution is a listing of the probabilities of all the possible outcomes that could result if the experiment is done.From the prior discussion we came to know thatthese are of the two types.

(a) Discrete → Discrete  probability is allowed to take any a limited number of values.

E.g. Binomial, Poisson variate.

(b) Continuous → the variable under consideration is allowed to take any value within given range.i.e. Normal variate

EXPECTED VALUE OF A RANDOM VARIABLE :

In a probabilistic situation, the selection of a strategy lead to a number of different possible outcomes, therefore, both average value and some measures of risk can be calculated by using the concept of expected value of the random variable.

Let  x be a random variable with respective probabilities

xj            :              x1                            x2                            x3    ………….    xn

p(xj)     :               p(x1)                      p(x2)                      p(x3) ………… p(xn)

The expected value of x, denoted by will be the ‘weighted average value’ of given variable :

= ∑   ( )

E(x) = ∑   ( )                                                 (x~discrete random variable)

( ) = ∫−∞∞   ( )                                                (x~continuous random variable)

Where f(x) is the probability density function

VARIANCE :

The expected value measures central tendency of a probability distribution,.Variance determines the dispersion’ or ‘variability’ to which the possible outcome/random values differ among themselves.

Var (X) = E(x –    )2

=  E (x2 – 2x   +   2)

=  E(x2) – 2   E(x) +   2

=  E(x2) –   2

As  E (x2)   =   ∑ =0 xj2 p(xj)

= ∑ =0                      p(xj)

Standard Deviation (x)

X    = √     ( ) = √∑ =1(  −   )2 ( 1)

The following are the important properties of an expected value of a random variable:

1.  The expected value of a constant c is constant. That is, E(c) = c, for every constant c.

2.  The expected value of the product of a constant c and a random variable x is equal to constant c times the expected value of the random variable. That is, E(cx) + cE(x)

3.  The expected value of a linear function of a random variable is same as the linear function of its expectation. That is, E(a + bx) = a + bE(x)

4.  The expected value of the product of two independent random variable is equal to the product of their individual expected values. That is, E(xy) = E(x) E(y).

5.  The expected value of the sum of the two independent random variables is equal to the sum of their individual expected values. That is, E(x ) + E(y) = E(x+y)

6.  The variance of the product of a constant and a random variable, x is equal to the constant squared times the variance of the random variable, x. That is, var(cx) = 2Var(x).

7.  The variance of the sum ( or difference) of two independent random variables equals the sum of their individual variance. That is, Var(x ± y) = Var(x) ± Var(y).

Example : An experiment consists of three independent tosses of a fair coin, let:

X = The number of heads, Y = the number of head runs, Z = The length of head runs, a head run being defined as consecutive occurrence of at least two heads, its length then being the number of heads occurring together in three tosses of the coin.

Find the probability function of (i) X, (ii) Y, (iii) Z, (iv) X + Y, and (v) XY and construct probability tables.

Solution

		X	Y	Z	X+Y	XY
1	HHH	3	1	3	4	3
2	HHT	2	1	2	3	2
3	HTH	2	0	0	2	0
4	HTT	1	0	0	1	0
5	THH	2	1	2	3	2
6	THT	1	0	0	1	0
7	TTH	1	0	0	1	0
8	TTT	0	0	0	0	0

Here sample space is :  S = { HHH, HHT, HTH, HTT, THH, THT, TTH,TTT}

(i) Obviously X is a random.variable. which can take the values 0,1,2 and 3. P(3) = P(HHH) =(1/2)3 = 18

P(2)= P(HHT ᴗ HTH ᴗ THH)

=  (      ) +   (      ) +   (      ) = 18 + 18 + 18 = 38 .

Similarly p(1) = 38 and p(0) = 18

These Probabilities could also be obtained directly from the Table 1

Table 2 : PROBBILITY DISTRIBUTION OF X

Q.1 A company introduces a new product in the market and expects to make a profit of Rs. 2.5 lakh during the first year if the demand is good; Rs. 1.5 lakh if the demand is ‘moderate’; and a loss of Rs. 1.0 lakh if the demand is “poor”. Market research studies indicate that the probabilities for the demand to be good and moderate are 0.2 and 0.5 respectively. Find the company’s expected profit and the standard deviation.

Solution : Let X1 be the random variable representing profit in 3 types of demand.

Thus, x may assume the values :

X1 = Rs. 2.5 lakh when demand is good.

X2 = Rs. 1.5 lakh when demand is Moderate

X3 = Rs. 1 lakh when demand is  poor.

Since the events are mutually exclusive and exhaustive .Therefore,

p(X1) + p(X2) + p(X3)=1

0.2 + 0.05 + p (X3)=1

p(X3) = 0.3

So, the expected profit is given by

E(x) = 2.5 x 0.2 + 1.5 x 0.5 + (–1) x (.3)

= Rs. 0.95 lakh

Also E (x2) = x12 p(x1) + x22 p(x2) + x32 p(x3)

= (2.5)2 x 0.2 + (1.5)2 (.5) + (–1)2 (.3)

= 2.675 lakh

=     √2.675 − (0.95)2

=  Rs. 1.331 lakh

Q.2 A doctor recommends to a patient to take a ‘paracetamol’ tablet for two weeks and there is equal chance for the patient to bring down this temperature between 20C and 40C. What is the average amount the patient is expected to bring down his temperature?

Solution :   It x is the random variable, then probability function is defined as

1

f(x) = { 2 , 2 <     < 4

0,   ℎ

The amount of temparature the patient is expected to lose :

Q.3 From a bag containing 3 red balls and 2 white balls, a man is to draw two balls at random without replacement. He gains Rs. 20 for each red ball and Rs. 10 for each white one. what is the expectation of his draw?

Solution : Let X be the random variable donating the number of red and white balls in a draw. So X can take values

∴ Expectation of the draw E(x) = 40 × 3/10 + 30 × 3 /150+ 20 × 1/10

= Rs. 32

(b) p (x = 1 red ball 1 white)  =

(c) p (x = 2 white balls)   =

3 1×  2 1

5 2   1

2 2

5 2

Q.4.An investor is contemplating to invest either in bonds that will pay 9% interest or in stocks. The stock prices may either go up by 20% or decrease by 10% or remain unchanged. The probabilities of these three occupancies we found by the investor as 0.3, 0.2 & 0.5 respectively. If the investor has initial capital of Rs. 1000, what should he do?

Solution : Let us assume that the handling costs remain same for both the donnatioves.

Thus the possible values of the stocks are :

1000 x 1.20 = 1200

1000 x 1.00 = 1000

1000 x 0.90 = 900

Hence, the expected monetary value of investing in stock

=  1200 x 0.3 + 1000 x 0.5 + 900 x 0.2

=  1040

Since the EMV of investing in stock lower (Rs. 1040), it is advisable for the investor to invest in bonds.

Q.5  A bag contains 2 white and 3 black balls. Four persons A, B, C, D in the order are named. Each draw one ball and do not replace it back. The person to draw a white ball receives Rs. 200. Determine their expectations.

Ans.     Since only 3 black balls are contained in the bag one person must win in the first attempt.

Probability that A wins or the ball drawn is white= 2/5.

A’s expectation =2/5 x 200 = Rs. 80/-

Probability that A looses and B wins =

B’s expectation =3/10*200   = Rs. 60/-

Probability that A and B looses but C wins =3/5-2/4-2/3/1/5

C’s expectation =1/5*200 = Rs. 40/-

D’s expectation = 1/10*200  = Rs. 20/-

Summary

In this module ,first random variable and types of random variables are explained .Then the concept of probability distribution, and its mean & standard distribution are discussed. This module is an extension of concept of probability to explain different probability distributions.As we see any given statistical experiment has more than one outcome. It is impossible to predict which of the outcome will occur if that experiment is performed. So decisions are made under uncertain conditions.In such type of situation probability distributions are useful to apply.